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A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then

a
T = t1+t2
b
T2 = t12+t22
c
T-1 = t2−1+t2−1
d
T−2 =t1−2+t2−2

detailed solution

Correct option is B

t1 = 2πmk1    and t2 = 2πmk2In series, effective spring constant, k = k1k2k1+k2∴  Time period, T = 2πm(k1+k2)k1k2----------(i)Now, t12+t22 = 4π2m(1k1+1k2) = 4π2m(k1+k2)k1k2             t12+t22 = T2.            [Using equation (ii)]

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